How to pass parameters to an eval based function inJavascript

Question

I am storing function body in string with function name.

function fnRandom(lim){
    var data=[];
    for(var i=0;i<lim;i++)
    {
        data=data.concat(Math.floor((Math.random() * 100) + 1));
    }
return data;
}

After selecting the functionName from a drop down I use eval to execute function body.

JSON.stringify(eval(this.selectedFunction.body));

I want to pass 'lim' to this execution or can I use functionName as initiating point for execution somehow?

Solution 1

Use Function constructor

var body = "console.log(arguments)"

var func = new Function( body );
func.call( null, 1, 2 ); //invoke the function using arguments

with named parameters

var body = "function( a, b ){ console.log(a, b) }"
var wrap = s => "{ return " + body + " };" //return the block having function expression
var func = new Function( wrap(body) );
func.call( null ).call( null, 1, 2  ); //invoke the function using arguments

Solution 2

Eval evaluates whatever you give it to and returns even a function.

var x = eval('(y)=>y+1');
x(3) // return 4

So you can use it like this:

var lim = 3;
var body = 'return lim+1;';
JSON.stringify(eval('(lim) => {' + body + '}')(lim)); //returns "4"

Another way using Function:

var lim = 3;
JSON.stringify((new Function('lim', this.selectedFunction.body))(lim));

Solution 3

You can use the Function object.

var param1 = 666;
var param2 = 'ABC';
var dynamicJS = 'return `Param1 = ${param1}, Param2 = ${param2}`';
var func = new Function('param1', 'param2', dynamicJS);
var result = func(param1, param2);
console.log(result);

Comments

[Barmar]

You didn't specify any arguments in the new Function() call.

[Barmar]

This only works if the function uses arguments instead of named parameters.

[Barmar]

he only has the function body, not the whole function.

[gurvinder372]

@Barmar yes, I have added a version with named parameters as well.

[Barmar]

Your function body is the whole function, not just the body.

[Barmar]

What if the parameter isn't named lim?

[Gerry]

He has to know how his parameters have to be named in his body-strings. Otherwise @gurvinder372 's answer will suit better.

[Barmar]

Yes, I think he has some fundamental design problems because of that. A solution like this doesn't work with arbitrary functions. He really needs to store more information in this.selectedFunction than just the body.

[gurvinder372]

If function is only body in OP's case, then named parameters cannot be used since we don't know the names of parameters used in the body

[Gerry]

I fully agree with you

[Barmar]

I know, I wrote that in a comment above. I think he has a design problem that makes this impossible to solve in a general way.